The slope of the normal to curve `y= x^(3) - 4x^(2)` at `(2 , -1)` is

Options

A. `(1)/(4)`
B. `(1)/(2)`
C. 4
D. `-4`

Answer

To find the slope of the normal to the curve \( y = x^3 - 4x^2 \) at the point \( (2, -1) \), we will follow these steps:

### Step 1: Find the derivative of the function
The first step is to find the derivative \( \frac{dy}{dx} \) of the function \( y = x^3 - 4x^2 \).

\[ \frac{dy}{dx} = \frac{d}{dx}(x^3) - \frac{d}{dx}(4x^2) \]

Using the power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \), we differentiate:

\[ \frac{dy}{dx} = 3x^2 - 8x \]

### Step 2: Evaluate the derivative at the point (2, -1)
Next, we evaluate the derivative at \( x = 2 \):

\[ \frac{dy}{dx} \bigg|_{x=2} = 3(2^2) - 8(2) \]

Calculating this gives:

\[ = 3(4) - 16 = 12 - 16 = -4 \]

### Step 3: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line at \( (2, -1) \) is \( -4 \), the slope of the normal line is:

\[ \text{slope of normal} = -\frac{1}{\text{slope of tangent}} = -\frac{1}{-4} = \frac{1}{4} \]

### Final Result
Thus, the slope of the normal to the curve at the point \( (2, -1) \) is:

\[ \frac{1}{4} \]

---